# 19. Properties of Triangles’ Solutions Part 2¶

1. The diagram is given below:

From figure, $\angle ADC = 90^\circ + B$

By applying $m:n$ rule in triangle $ABC,$ we get

$(1 + 1)\cot(90^\circ + B) = 1.\cot90^\circ - 1.\cot(A-90^\circ)$

$-2\tan B = 0 - \cot[-(90^\circ - A)]$

$-2\tan B = \tan A \Rightarrow \tan A + 2\tan B = 0$

1. Given, $\cot A + \cot B + \cot C = \sqrt{3}$

Squaring, $\cot^2A + \cot^2B + \cot^2C + 2\cot A\cot B + 2\cot B\cot C + 2\cot C\cot A = 3$

Since $A + B + C = \pi \Rightarrow A + B = \pi - C$

$\cot(A + B) = \cot(\pi - C)$

$\frac{\cot A\cot B - 1}{\cot A + \cot B} = -\cot C$

$\cot A\cot B + \cot B\cot C + \cot C\cot A = 1$

$\Rightarrow \cot^2A + \cot^2B + \cot^2C + 2\cot A\cot B + 2\cot B\cot C + 2\cot C\cot A = 3(\cot A\cot B + \cot B\cot C + \cot C\cot A)$

$\frac{1}{2}[(\cot A - \cot B)^2 + (\cot B - \cot C)^2 + (\cot C - \cot A)^2] = 0$

$\Rightarrow \cot A = \cot B = \cot C$

$\Rightarrow A = B = C$ i.e. triangle is equilateral.

2. Given, $(a^2 + b^2)\sin(A - B) = (a^2 - b^2)\sin(A + B)$

$(\sin^2A + \sin^2B)\sin(A - B) = (\sin^2A - \sin^2B)\sin(A + B)$

$[\because \sin^2A - \sin^2B = \sin(A + B)\sin(A - B)]$

$\Rightarrow (\sin^2A + \sin^2B)\sin(A - B) = \sin^2(A + B)\sin(A - B)$

$\Rightarrow \sin(A - B)[\sin^2A + \sin^2B - \sin^2C] = 0$

Either $\sin(A - B) = 0$ or $\sin^2A + \sin^2B - \sin^2C = 0$

$A = B$ or $a^2 + b^2 - c^2 = 0$

Thus, triangle is either isosceles or right angled.

3. R.H.S. $= c(\cos A\cos\theta + \sin A\sin\theta) + a(\cos C\cos\theta - \sin C\sin\theta)$

$= \cos\theta(c\cos A + a\cos C) + \sin\theta(c\sin A - a\sin C)$

$=b\cos\theta + \sin\theta\left(c.\frac{a}{2R} - a.\frac{c}{2R}\right)$

$= b\cos\theta =$ L.H.S.

4. The diagram is given below:

$\frac{b}{\sin(B + \theta)} = \frac{a}{2\sin(A - \theta)}$

$\frac{c}{\sin[\pi -(B + \theta)]} = \frac{a}{2\sin\theta}$

$\Rightarrow b\sin(A - \theta) = c\sin\theta$

$\Rightarrow \sin B(\sin A\cos\theta - \cos A\sin\theta) = \sin C\sin\theta = (\sin A\cos B + \sin B \cos A)\sin\theta$

$\Rightarrow \cot\theta = 2\cot A + \cot B$

5. The diagram is given below:

$\Delta ABC = \Delta ABD + \Delta ACD$

$\frac{1}{2}bc\sin A = \frac{1}{2}AD.c\sin\frac{A}{2} + \frac{1}{2}AD.b\sin\frac{A}{2}$

$\Rightarrow AD = \frac{2bc}{b + c}\cos\frac{A}{2}$

6. The diagram is given below:

$\angle ACB = \pi - (\alpha + \beta + \gamma)$

$\sin ACB = \sin(\alpha + \beta + \gamma)$

In $\triangle ABC$

$\frac{AB}{\sin ACB} = \frac{AC}{\sin\gamma}$

$AC = \frac{a\sin\gamma}{\sin(\alpha + \beta + \gamma)}$

In $\triangle ACD$

$\frac{CD}{\sin\alpha} = \frac{AC}{\sin\beta}$

$CD = \frac{a\sin\alpha\sin\gamma}{\sin\beta\sin(\alpha + \beta + \gamma)}$

7. Given, $2\cos A = \frac{\sin B}{\sin C}$

$2\cos A\sin C = \sin B = \sin[\pi - (A + C)] = \sin(A + C)$

$2\cos A\sin C = \sin A\cos C + \cos A\sin C$

$\cos A\sin C = \sin A\cos C\Rightarrow \tan A = \tan C$

$\Rightarrow A = C$

Thus, the triangle is isosceles.

8. Let such angles be $A$ and $B.$ Then,

$\cos A = \frac{1}{a}$ and $\cos B = \frac{1}{b}$

$\Rightarrow \sin A\cos A = \sin B\cos B$

$\sin 2A = \sin 2B$ or $\sin 2A = \sin[\pi - 2B]$

$A = B$ or $A + B = \frac{\pi}{2} \Rightarrow C = \frac{\pi}{2}$

Thus, the triangle is either isosceles or right angled.

9. Given $a\tan A + b\tan B = (a + b)\tan \frac{A + B}{2}$

$\Rightarrow a\left(\frac{\sin A}{\cos A} - \frac{\sin \frac{A + B}{2}}{\cos\frac{A + B}{2}}\right) = b\left(\frac{\sin\frac{A + B}{2}}{\cos \frac{A + B}{2}} - \frac{\sin B}{\cos B}\right)$

$\Rightarrow a.\frac{\sin\frac{A - B}{2}}{\cos A\cos \frac{A + B}{2}} = b.\frac{\sin\frac{A - B}{2}}{\cos B\cos\frac{A + B}{2}}$

$\Rightarrow \tan A = \tan B \Rightarrow A = B$

Thus, the triangle is isosceles.

10. Given, $\frac{\tan A - \tan B}{\tan A + \tan B} = \frac{c - b}{c}$

$\Rightarrow \frac{\frac{\sin A}{\cos A} - \frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}} = \frac{\sin C - \sin B}{\sin C}$

$\Rightarrow \frac{\sin A\cos B - \sin B\cos A}{\sin A\cos B + \sin B\cos A} = \frac{\sin (A + B) - \sin B}{\sin (A + B)}$

$\Rightarrow \frac{\sin A\cos B - \sin B\cos A}{\sin(A + B)} = \frac{\sin(A + B) - \sin B}{\sin(A + B)}$

$\Rightarrow \sin A\cos B - \sin B\cos A = \sin A\cos B + \sin B\cos A - \sin B$

$\Rightarrow \sin B = 2\sin B\cos A \Rightarrow \cos A = \frac{1}{2} \Rightarrow A = 60^\circ.$

11. We know that $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$

Given, $c^4 - 2(a^2 + b^2)c^2 + a^4 + a^2b^2 + b^4 = 0$

$\Rightarrow a^4 + b^4 + c^4 - 2a^2c^2 - 2b^2c^2 + 2a^2b^2 = a^2b^2$

$(a^2 + b^2 -c^2)^2 = a^2b^2 \Rightarrow a^2 + b^2 + c^2 = \pm ab$

$\Rightarrow \cos C = \pm\frac{1}{2} \Rightarrow A = 60^\circ$ or $120^\circ$

12. Given, $\frac{\cos A + 2\cos C}{\cos A + 2\cos B} = \frac{\sin B}{\sin C}$

$\Rightarrow \cos A(\sin C - \sin B) = 2\sin B\cos B - 2\sin C\cos C = \sin 2B - \sin 2C$

$\Rightarrow 2\cos A.\cos\frac{B + C}{2}\sin\frac{C - B}{2} = 2\cos(B + C)\sin(B - C)$

$\cos A.\cos\frac{B + C}{2}\sin\frac{C - B}{2} = -2\cos A.\sin\frac{B - C}{2}.\cos\frac{B - C}{2}$

If $B = C$ then above it $0 = 0$ i.e. triangle is isosceles.

If $A = 90^\circ$ then above is $0 =0$ i.e. triangle is right angled.

13. $\because \tan\frac{A}{2}, \tan\frac{B}{2}, \tan\frac{C}{2}$ are in A.P.

$\tan\frac{A}{2} - \tan\frac{B}{2} = \tan\frac{B}{2} - \tan \frac{C}{2}$

$\frac{\sin\left(\frac{A}{2} - \frac{B}{2}\right)}{\cos\frac{A}{2}\cos\frac{B}{2}} = \frac{\sin\left(\frac{B}{2} - \frac{C}{2}\right)}{\cos\frac{B}{2}\cos\frac{C}{2}}$

$\sin\left(\frac{A}{2} - \frac{B}{2}\right)\cos\frac{C}{2} = \sin\left(\frac{B}{2} - \frac{C}{2}\right)\cos\frac{A}{2}$

$\sin\left(\frac{A}{2} - \frac{B}{2}\right)\sin\left(\frac{A}{2} + \frac{B}{2}\right) = \sin\left(\frac{B}{2} - \frac{C}{2}\right)\sin\left(\frac{B}{2} + \frac{C}{2}\right)$

$\Rightarrow \cos B - \cos A = \cos C - \cos B$

Thus, $\cos A, \cos B, \cos C$ are in A.P.

14. Given, $a\cos^2\frac{C}{2} + c\cos^2\frac{A}{2} = \frac{3b}{2}$

$a.\frac{s(s - c)}{ab} + c.\frac{s(s - a)}{bc} = \frac{3b}{2}$

$\frac{s}{b}[2s - a - c] = \frac{3b}{2}$

$2s = 3b \Rightarrow a + c = 2b$

We have to prove that $\cot\frac{A}{2} + \cot\frac{C}{2} = 2\cot\frac{B}{2}$

L.H.S. $= \frac{s(s - a)}{\Delta} + \frac{s(s - c)}{\Delta}$

$= \frac{s}{\Delta}(2s - a - c) = \frac{2s(s - b)}{\Delta} = 2\cot\frac{B}{2} =$ R.H.S.

15. Given, $a^2, b^2, c^2$ are in A.P.

$\Rightarrow b^2 - a^2 = c^2 - b^2$

$\Rightarrow \sin^2B - \sin^2A = \sin^2C - \sin^2B$

$\Rightarrow \sin(B + A)\sin(B - A) = \sin(C + B)\sin(C - B)$

$\Rightarrow \sin C\sin(B - A) = \sin A\sin(C - B)$

$\Rightarrow \frac{\sin A\cos B - \cos B\sin A}{\sin A\sin B} = \frac{\sin B\cos C - \sin C\cos B}{\sin B\sin C}$

$\Rightarrow \cot B - \cot A = \cot C - \cot B$

$\therefore \cot A, \cot B, \cot C$ are in A.P.

16. Since $A, B, C$ are in A.P. $\Rightarrow 2B = A + C \Rightarrow A + B + C = 3B = 180^\circ \Rightarrow B = 60^\circ$

Given, $2b^2 = 3c^2$

$2.\sin^2B = 3.\sin^2C \Rightarrow \sin C = \pm\frac{1}{\sqrt{2}}$

$\sin C\neq -\frac{1}{\sqrt{2}}$ because $C < 120^\circ$

$\sin C = \frac{1}{\sqrt{2}} \Rightarrow C = 45^\circ$

$\Rightarrow A = 75^\circ$

17. Given, $\tan\frac{A}{2}, \tan\frac{B}{2}, \tan\frac{C}{2}$ are in H.P.

$\Rightarrow \cot \frac{A}{2}, \cot\frac{B}{2}, \cot\frac{C}{2}$ are in A.P.

$\cot\frac{B}{2} - \cot\frac{A}{2} = \cot\frac{C}{2}- \cot\frac{B}{2}$

$\frac{s(s - b) - s(s - a)}{\Delta} = \frac{s(s - c) - s(s - b)}{\Delta}$

$a - b = b - c$

$a, b, c$ are in A.P.

18. Given, $\frac{\sin A}{\sin C} = \frac{\sin(A - B)}{\sin(B - C)}$

$\frac{\sin A}{\sin C} = \frac{\sin A\cos B - \sin B \cos A}{\sin B\cos C - \sin C\cos B}$

$\sin A\sin C\cos C + \sin B\sin C\cos A = 2\sin A\sin C\cos B$

$\sin B\sin(A + C) = 2\sin A\sin C\cos B$

$\sin^2B = 2\sin A\sin C\cos B$

$\cos B = \frac{b^2}{2ac} = \frac{a^2 + c^2 - b^2}{2ac}$

$\Rightarrow c^2 + a^2 = 2b^2$

Thus, $a^2, b^2, c^2$ are in A.P.

19. Given, $2\sin B = \sin A + \sin C$

$4\sin\frac{B}{2}\cos\frac{B}{2} = 2\sin\frac{A + C}{2}\cos\frac{A - C}{2} = 2\cos\frac{B}{2}\cos\frac{A - C}{2}$

$2\sin\frac{B}{2} = 3\cos\frac{A + C}{2} = \cos\frac{A - C}{2}$

$3\sin\frac{A}{2}\sin\frac{C}{2} = \cos\frac{A}{2}\cos\frac{C}{2}$

$3\tan\frac{A}{2}\tan\frac{C}{2} = 1$

20. Given, $a^2, b^2, c^2$ are in A.P.

$b^2 - a^2 = c^2 - b^2$

$\sin^2B - \sin^2A = \sin^2C - \sin^2B$

$\sin(A + B)\sin(A - B) = \sin(B + C)\sin(B - C)$

$\sin C\sin(A - B) = \sin A\sin(B - C)$

$\cos B -\cot A\sin B = \sin B\cot C - \cos B$

$2\cos B = \sin B(\cot A + \cot C)$

$2\cot B = \cot A + \cot C$

$\therefore \cot A, \cot B, \cot C$ are in A.P.

$\therefore \tan A,\tan B, \tan C$ are in H.P.

21. We have proven in previous problem that $\therefore \cot A, \cot B, \cot C$ are in A.P.

22. Since $A, B, C$ are in A.P. $\Rightarrow 2B = A + C \Rightarrow A + B + C = 3B = 180^\circ \Rightarrow B = 60^\circ$

$b:c = \sqrt{3}:\sqrt{2} \Rightarrow \sin C = \frac{\sqrt{3}}{2}.\sqrt{\frac{2}{3}} = \frac{1}{\sqrt{2}}$

$\Rightarrow C = 45^\circ \Rightarrow A = 75^\circ$

23. Let the sides are $a, b, c$ then $2b = a + c.$ Also, let $a$ to be greatest and $c$ to be smallest side.

Then, $A = 90 + C$ then $90 + C + B + C = 180 \Rightarrow B = 90 -2C$

$\frac{a}{\sin(90 + C)} = \frac{b}{\sin(90 - 2C)} = \frac{c}{\sin C} = 2R$

$4R\cos2C = 2R\cos C + 2R\sin C$

$2\cos 2C = \cos C + \sin C$

Squaring, we get

$4(1 - \sin^22C) = 1 + \sin 2C \Rightarrow \sin2C = \frac{3}{4}$ when $1 + \sin 2C \neq 0$

When $1 + \sin 2C = 0 \Rightarrow C = \frac{3\pi}{4}$ which is not possible.

$\because \sin 2C = \frac{3}{4} \Rightarrow \cos 2C = \frac{\sqrt{7}}{4}$

Now $\sin C$ and $\cos C$ can be found and ratio can be evaluated.

24. $\because a, b, c$ are in A.P. $2b = a + c \Rightarrow a = 2b - c$

$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{b^2 + c^2 -4b^2 -c^2 +4bc}{2bc}$

$= \frac{4bc - 3b^2}{2bc} =\frac{4c - 3b}{2c}$

25. The diagram is given below:

Let $AB = 2, AD = 5, BC = 3$ annd $CD=x$

Since it is cyclic quadrilateral $\angle C = 120^\circ$

Applying cosine rule in $\Delta ABD,$ we have

$\cos 60^\circ = \frac{AB^2 + AD^2 - BD^2}{2.AB.AD} \Rightarrow BD^2 = 19$

Applying cosine rule in $\Delta BCD,$ we have

$\cos 120^\circ = \frac{BC^2 + CD^2 - BD^2}{2.BC.BD} \Rightarrow x^2 + 3x -10 = 0$

$x = -5, 2$ but $x$ cannot be -ve. $\therefore x = 2$

26. Given $(a + b + c)(b + c - a) = 3bc$

$b^2 + c^2 - a^2 + 2bc = 3bc$

$\frac{b^2 + c^2 - a^2}{2bc} = \frac{1}{2}$

$\cos A = \cos 60^\circ$

$A = 60^\circ$

27. Since $AD$ is the median

$\therefore AB^2 + AC^2 = 2BD^2 + 2AD^2$

$\Rightarrow b^2 + c^2 = \frac{a^2}{4} + 2AD^2$

$4AD^2 = b^2 + c^2 + (b^2 + c^2 - a^2)$

$\cos A = \frac{1}{2} = \frac{b^2 + c^2 - a^2}{2bc}$

$\Rightarrow 4AD^2 = b^2 + c^2 + bc$

28. The diagram is given below:

Since $AD$ is the median

$\therefore AB^2 + AC^2 = 2BD^2 + 2AD^2$

$\Rightarrow AD^2 = \frac{2b^2 + 2c^2 - a^2}{4}$

$AO = \frac{2}{3}AD = \frac{2}{3}.\frac{1}{2}\sqrt{2b^2 + 2c^2 - a^2}$

Similalry $BO = \frac{1}{3}\sqrt{2c^2 + 2a^2 - b^2}$

$\because \angle AOB = 90^\circ$

$\therefore AO^2 + BO^2 = AB^2$

$\Rightarrow a^2 + b^2 = 5c^2$

29. Given, $\frac{\tan A}{1} = \frac{\tan B}{2} = \frac{\tan C}{3} = k$

Since $A, B, C$ are the angles of a triangle

$\therefore \tan A + \tan B + \tan C = \tan A\tan B\tan C$

$k + 2k + 3k = k.2k.3k \Rightarrow k = 1$ as if $k = -1$ sum of angles will be greater than $180^\circ.$

$\tan A = 1 \Rightarrow \sin A = \frac{1}{\sqrt{2}}$

$\tan A = 2 \Rightarrow \sin A = \frac{2}{\sqrt{5}}$

$\tan A = 3 \Rightarrow \sin A = \frac{3}{\sqrt{10}}$

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

$\sqrt{2}a = \frac{\sqrt{5}b}{2} = \frac{\sqrt{10}c}{3}$

$6\sqrt{2}a = 3\sqrt{5}b = 2\sqrt{10}c$

30. For a triangle sides are positive i.e. $a > 0, b > 0, c >0$ where $a,b,c$ are the sides.

$2x + 1 > 0 \Rightarrow x > -\frac{1}{2}$

$x^2 - 1>0 \Rightarrow x > 1$ because side cannot be negative.

$x^2 + x + 1 > 0~\forall x>1$

$a - b = x(x - 1) > 0 \Rightarrow a > b$

$a - c = x + 2 > 0 \Rightarrow a > c$

Hence $a$ is the greatest side.

$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{(2x + 1)^2 + (x^2 - 1)^2 - (x^2 + x + 1)}{2(2x + 1)(x^2 - 1)}$

$= -\frac{1}{2}$

$\Rightarrow A = 120^\circ$

31. Let the sides be $x, x+1, x+2$ where $x > 0$ and is a natural number. Let the smallest angle be $\theta$

$\angle C = \theta \therefore \angle A = 2\theta$

Applying sine law

$\frac{x}{\sin\theta} = \frac{x + 1}{\sin(\pi - 3\theta)} = \frac{x + 2}{\sin2\theta}$

$\Rightarrow \frac{x}{\sin\theta} = \frac{x + 2}{\sin2\theta} \Rightarrow 2\cos\theta = \frac{x + 2}{x}$

$\Rightarrow \frac{x}{\sin\theta} = \frac{x + 1}{\sin3\theta} = \frac{x + 1}{3\sin\theta - 4\sin^2\theta}$

$\Rightarrow 3 - 4\sin^2\theta = \frac{x + 1}{x}$

$\Rightarrow 4\cos^2\theta = \frac{2x + 1}{x} = \frac{(x + 2)^2}{x^2}$

$\Rightarrow x^2 - 3x - 4 = 0$

$x = 4, -1$ but $-1$ is not a natural number so $x = 4.$ Hence sides are $4,5,6.$

32. Given, $a = 6$ cm, $\Delta = 12$ sq. cm. and $B - C = 60^\circ$

$\Delta = \frac{1}{2}ab\sin C = \frac{1}{2}a.k\sin B\sin C$

$= \frac{1}{2}.a.\frac{a}{\sin A}\sin B\sin C$

$\Delta = \frac{1}{2}a^2\sin B\sin C = \frac{18\sin B\sin C}{\sin A}$

$\Rightarrow \frac{2}{3} = \frac{2\sin B\sin C}{2\sin A} = \frac{\cos(B - C) - \cos(B + C)}{2\sin A}$

$\Rightarrow \frac{2}{3} = \frac{\cos60^\circ - \cos(\pi - A)}{2\sin A}$

$\Rightarrow 8\sin A - 6\cos A = 3$

33. Given, $\cos\theta = \frac{a}{b + c} \Rightarrow 1 + \cos\theta = \frac{a + b + c}{b + c}$

$\Rightarrow 2\cos^2\frac{\theta}{2} = \frac{a + b + c}{b + c}$

$\sec^2\frac{\theta}{2} = \frac{2(b + c)}{a + b + c}$

$1 + \tan^2\frac{\theta}{2} = \frac{2(b + c)}{a + b + c}$

Similarly, $1 + \tan^2\frac{\phi}{2} = \frac{2(c + a)}{a + b + c}$

and $1 + \tan^2\frac{\psi}{2} = \frac{2(a + b)}{a + b + c}$

Adding, we get $3 + \tan^2\frac{\theta}{2} + \tan^2\frac{\phi}{2} + \tan^2\frac{\psi}{2} = \frac{4(a + b + c)}{a + b + c}$

$\Rightarrow \tan^2\frac{\theta}{2} + \tan^2\frac{\phi}{2} + \tan^2\frac{\psi}{2} = 1$

34. Since $C$ is the angle of a triangle, $\sin C\leq 1$

$\therefore \cos A\cos B + \sin A\sin B\geq \cos A\cos B + \sin A\sin B\sin C$

$\Rightarrow \cos(A - B)\geq 1$

But $cos(A - B)$ cannot be greater than $1. \therefore \cos(A - B) = 1\Rightarrow A = B$

Now, $\cos A\cos B + \sin A\sin B\sin C = 1$

$\Rightarrow \cos^2A + \sin^2A\sin C = 1$

$\Rightarrow \sin C= 1 \Rightarrow C=90^\circ\Rightarrow A = B = 45^\circ$

$\Rightarrow a:b:c = \sin A:\sin B: \sin C = 1:1:\sqrt{2}$

35. From the question, $\sin A\sin B\sin C = p$ and $\cos A\cos B\cos C = q$

$\therefore \tan A\tan B\tan C = \frac{p}{q}$

Since we are dealing with a triangle $\therefore \tan A + \tan B + \tan C = \tan A\tan B\tan C$

$\Rightarrow \tan A + \tan B + \tan C = \frac{p}{1}$

Now, $\tan A\tan B + \tan B\tan C + \tan C\tan A = \frac{\sin A\sin B\cos C + \sin B\sin C\cos A + \sin A\sin C\cos B}{\cos A\cos B\cos C}$

[ We know that in a triangle $2\sin A\sin B\cos C = \sin^2A + \sin^2B - \sin^2C$ ]

$\Rightarrow \frac{1}{2q}[(\sin^2A + \sin^2B - \sin^2C) + (\sin^2B + \sin^2C - \sin^2A) + (\sin^2C + \sin^2A - \sin^2B)]$

$= \frac{1}{2q}[\sin^2 + \sin^2B + \sin^2C]$

$= \frac{1}{2q}\left[\frac{(1 - \cos2A) + (1 - \cos2B) + (1 - \cos2C)}{2}\right]$

$= \frac{1}{4q}[4 + 4\cos A\cos B\cos C] = \frac{1 + q}{q}$

Thus, we see that $\tan A, \tan B, \tan C$ are roots of the given equation.

36. Given, $\sin^3\theta = \sin(A - \theta)\sin(B - \theta)\sin(C - \theta)$

$4\sin^3\theta = 2\sin(A - \theta)[2\sin(B - \theta)\sin(C - \theta)]$

$= 2\sin(A - \theta)[\cos(B - C) - \cos(B + C - 2\theta)]$

$= 2\sin(A - \theta)\cos(B - C) -2\sin(A - \theta)\cos(B + C - 2\theta)$

$= \sin(A + B - C - \theta) + \sin(A + C - \theta - B) - \sin(A + B + C - 3\theta) + \sin(\pi - 2B - \theta)$

$\sin 3\theta + 4\sin^3\theta = \sin(2A + \theta) + \sin(2B + \theta) + \sin(2C + \theta)$

$3\sin\theta = (\sin2A + \sin2B + \sin2C)\cos\theta + (\cos2A + \cos2B + \cos2C)\sin\theta$

[ $\because \sin2A + \sin2B + \sin 2C = 4\sin A\sin B\sin C$ when $A + B + C = \pi$ ]

$(1 - \cos 2A) + (1 - \cos 2B) + (1 - \cos 2C)\sin\theta = 4\sin A\sin B\sin C.\cos\theta$

$2[(\sin^2A + \sin^2B + \sin^2C)\sin\theta] = 4\sin A\sin B\sin C\cos\theta$

$2\sin\theta[(\sin^2A + \sin^2B - \sin^2C) + (\sin^2B + \sin^2C - \sin^2A) + (\sin^2C + \sin^2A - \sin^2B)] = 4\sin A\sin B\sin C\cos\theta$

[ $\because \sin^2 + \sin^2B - \sin^2C = 2\sin A\sin B\cos C$ ]

$\Rightarrow \cot\theta = \cot A + \cot B + \cot C$

37. From question $\frac{b}{c} = r \therefore b = cr$

Let $AD\perp BC$ and let $AD = h$

We have to prove that $h\leq \frac{ar}{1 - r^2}$

$\Delta ABC = \frac{1}{2}c.cr.\sin A = \frac{1}{2}ah$

$h = \frac{c^2r\sin A}{a}$

$\cos A = \frac{c^2 + c^2r^2 - a^2}{2.c.cr}$

$c^2 = \frac{a^2}{1 + r^2 - 2r\cos A}$

$\Rightarrow h = \frac{a^2r\sin C}{a(1 + r^2 - 2r\cos a)} = \frac{ar\sin A}{1 + r^2 - 2r\cos A}$

$= \frac{ar.\frac{2\tan\frac{A}{2}}{1 + \tan^2\frac{A}{2}}}{1 + r^2 - 2r\frac{1 - \tan^2\frac{A}{2}}{1 + \tan^2\frac{A}{2}}}$

$\Rightarrow (1 + r^2)\tan^2\frac{A}{2} -\frac{2ar}{h}\tan\frac{A}{2} + (1 - r)^2 = 0$

This is a quadratic equation in $\tan\frac{A}{2}$ and since it will be read $D \geq 0$

$\Rightarrow \frac{4a^2r^2}{h^2} - 4(1 + r)^2(1 - r)^2\geq 0$

$h \leq \frac{ar}{1 - r^2}$

38. Given $b.c = k^2,$ now

$\cos A = \frac{b^2 + c^2 - a^2}{2bc} \Rightarrow 2k^2\cos A = b^2 + \frac{k^4}{b^2} - a^2$

$b^4 - (a^2 + 2k^2\cos A)b^2 + k^4 = 0$ which is a quadratic equation in $b^2.$

The triangle will not exists if discriminant is less than zero for above equation because then $b$ will become a complex number.

$\Rightarrow (a^2 + 2k^2\cos A)^2 - 4k^4 < 0$

$\Rightarrow [a^2 + 2k^2(1 + \cos A)][a^2 - 2k^2(1 - \cos A)] < 0$

$\Rightarrow \left(a^2 + 4k^2\cos^2\frac{A}{2}\right)\left(a^2 - 4k^2\sin^2\frac{A}{3}\right) < 0$

$\Rightarrow a^2 - 4k^2\sin^2\frac{A}{2} , 0 \left[\because a^2 + 2k^2\cos^2\frac{A}{2} > 0\right]$

$\Rightarrow \left(a + 2k\sin\frac{A}{2}\right)\left(a - 2k\sin\frac{A}{2}\right) < 0$

$\Rightarrow a - 2k\sin\frac{A}{2} < 0 \left[\because] a + 2k\sin\frac{A}{2} > 0\right]$

$\Rightarrow a < 2k\sin\frac{A}{2}$

39. The diagram is given below:

The diagram is a top view. Let $O$ be the top point and $O'$ the center of ring which is $12$ cm below $O$ in the diagram(not shown).

In triangle $OO'A, AO' = 5$ cm, $OO' = 12$ cm

$AO = \sqrt{12^2 + 5^2} = 13$ cm

Now sides of a regular hexagon are equal to the circumscribing circle. $AB= 5$ cm.

$\cos AOB = \frac{13^2 + 13^2 - 5^2}{2.13.13} = \frac{313}{338}$

40. Given, $2b = 3a$ and $\tan^2\frac{A}{2}= \frac{3}{5}$

$\cos A = \frac{1}{\sqrt{1 + \tan^2\frac{A}{2}}} = \frac{1}{\sqrt{1 + \frac{3}{5}}} = \sqrt{\frac{5}{8}}$